Mon, 9 Jun 2014 15:43:46 +0200
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> ("Lars Hellström"'s message of "Mon, 9 Jun 2014 14:25:07
Lars Hellström <[log in to unmask]> writes:
> Oh yes. One of the really big mistakes people make when trying to
> implement mathematics is believing that the mathematical formula
> language is consistent, just because it is precise. Juxtaposition can
> denote pretty much *anything* (depending on context), and because
> juxtaposition is multiplication, it follows that pretty much anything
> can be regarded as a kind of multiplication. ;-)
More like "anything can be regarded as a kind of linear operator with
respect to the following expression".
d/dx is not really something you "multiply" with. Everybody is
comfortable writing and reading x y, but when it is x x, it looks
When you take something like Fourier transforms, linear time-invariant
operators _are_ a multiplication after transform.
If you take the t->f kind of transform electrical engineers prefer
(because you don't need to fiddle with asymmetric or weird scaling
factors), you can basically say
\nabla -> 2\pi \vec f
Which makes little sense on the surface, but a rot written as
\nabla \times transforms to 2\pi \vec f \times, a div written as
\nabla \cdot transforms to 2\pi \vec f \cdot, a gradient written
as \nabla \phi transforms to 2\pi \vec f \Phi, so for a number of
purposes, operator juxtaposition works in a rather multiplication-like
Of course, like with other mathematical jigglery-pokery in engineering
professions, you'll not find anything in the books used to teach the
engineers, and the books used to teach mathematicians use completely
other terminology and conventions not convenient for the kind of vector
analysis an engineer might want to do.
And the theoretical physicists do something else entirely in order not
to get caught in the crosswinds.
You'll be returning to your regular scheduled topic of LaTeX after the